Solution of EECS 300 Test 6 F08
1.
() (
)
()
A current i t = 3t + 4 A is flowing through a 10 mF capacitor and a 2H inductor in series. The voltage v C t
across the capacitor at time t = 0 is 1200 V. Find the numerical values of the capacitor voltage and the inductor
voltage at time t = 5 s .
i (t)
10 mF
v (t)
C
()
v C t = 1200V +
t
t
1
i
d
=
1200
+
100
3 + 4 d V
C
C0
0
()
(
)
()
t
v C t = 1200 + 100 3 2 / 2 + 4 V
0
() (
)
v C t = 1200 + 300t 2 / 2 + 400t V
() (
( ))
( )
v C 5 = 1200 + 300 25 / 2 + 400 5 V = 6950V
()
vL t = L
di
= 2H 3 A/s = 6V v L 5 = 6V
dt
( )(
)
()
2H
v (t)
L
2.
In the circuit below, the initial capacitor voltage at time t = 0 is 24V. The energy decays toward zero as
time passes. How much energy will be dissipated in the resistor over the time range 0 < t < ?
50μF
10Ω
All the initial energy in the capacitor must be dissipated in the resistor. The initial energy stored in the
capacitor is
(
)
(
)
(
EC = 1 / 2 Cv 2 = 1 / 2 50 μ F 24V
3.
)
2
= 14.4mJ
The voltage source in the circuit below is a constant 1V and the current source is a constant 1A. Find the
numerical values of the voltages v1 ,v2 and v3 .
1Ω
v
3
1F
1Ω
1H
1V
1Ω
v1
1F
1H
v2
1F
1Ω
1A
We can simplify the circuit for this problem by replacing all capacitors by open circuits and all inductors
by short circuits. Also the upper-right hand resistor is shorted out and can be removed without changing
any voltages or currents.
1Ω
v
3
1Ω
1V
v
1
v
2
1Ω
1A
Due to the 1V source acting alone:
Current source is open-circuited. Current flows through the middle resistor path and down through the
right hand resistor. Half of the 1V source voltage appears across each of the two resistors in that path.
v1 = 1V , v2 = 0.5 V and v3 = 0.5V
Due to the 1A source acting alone:
Voltage source is short-circuited. The source current splits between the right-hand resistor and the middle
resistor, 1/2A going each way. v1 is zero because it is across a short circuit. v2 is 0.5 V and v3 is 0.5 V.
Including both sources:
v1 = 1V , v2 = 1 V and v3 = 0
Solution of EECS 300 Test 6 F08
1.
() (
)
()
A current i t = 5t + 2 A is flowing through a 10 mF capacitor and a 2H inductor in series. The voltage v C t
across the capacitor at time t = 0 is 1000 V. Find the numerical values of the capacitor voltage and the inductor
voltage at time t = 5 s .
i (t)
10 mF
v (t)
C
()
v C t = 1000V +
t
t
1
iC d = 1000 + 100 5 + 2 d V
C0
0
()
(
)
()
t
v C t = 1000 + 100 5 2 / 2 + 2 V
0
() (
)
v C t = 1000 + 500t 2 / 2 + 200t V
() (
( ))
( )
v C 5 = 1000 + 500 25 / 2 + 200 5 V = 8250V
()
vL t = L
di
= 2H 5 A/s = 10V v L 5 = 10V
dt
( )(
)
()
2H
v (t)
L
2.
In the circuit below, the initial capacitor voltage at time t = 0 is 16V. The energy decays toward zero as
time passes. How much energy will be dissipated in the resistor over the time range 0 < t < ?
50μF
10Ω
All the initial energy in the capacitor must be dissipated in the resistor. The initial energy stored in the
capacitor is
(
)
(
)
(
EC = 1 / 2 Cv 2 = 1 / 2 50 μ F 16V
3.
)
2
= 6.4mJ
The voltage source in the circuit below is a constant 1V and the current source is a constant 1A. Find the
numerical values of the voltages v1 ,v2 and v3 .
1Ω
v
3
1F
1Ω
1H
1V
1Ω
v1
1F
1H
v2
1F
1Ω
1A
We can simplify the circuit for this problem by replacing all capacitors by open circuits and all inductors
by short circuits. Also the upper-right hand resistor is shorted out and can be removed without changing
any voltages or currents.
1Ω
v
3
1Ω
1V
v
1
v
2
1Ω
1A
Due to the 1V source acting alone:
Current source is open-circuited. Current flows through the middle resistor path and down through the
right hand resistor. Half of the 1V source voltage appears across each of the two resistors in that path.
v1 = 1V , v2 = 0.5 V and v3 = 0.5V
Due to the 1A source acting alone:
Voltage source is short-circuited. The source current splits between the right-hand resistor and the middle
resistor, 1/2A going each way. v1 is zero because it is across a short circuit. v2 is 0.5 V and v3 is 0.5 V.
Including both sources:
v1 = 1V , v2 = 1 V and v3 = 0